sunshine022uv
2022-01-09
Answered

Consider the following unbalanced equation: $C{a}_{3}{\left(P{O}_{4}\right)}_{2}\left(s\right)+3{H}_{2}S{O}_{4}\left(aq\right)\Rightarrow 3CaS{O}_{4}\left(s\right)+2{H}_{3}P{O}_{4}\left(aq\right)$ . What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 kg of calcium phosphate with 1.0 kg of concentrated sulfuric acid $(98\mathrm{\%}{H}_{2}S{O}_{4}$ by mass)?

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Cassandra Ramirez

Answered 2022-01-10
Author has **30** answers

Balance the equation:

$C{a}_{3}{\left(P{O}_{4}\right)}_{2\left(s\right)}+{3}_{H2}S{O}_{4\left(aq\right)}\Rightarrow 3CaS{O}_{4\left(s\right)}+{2}_{H3PO4\left(aq\right)}$

Solve for the masses of the reactants:

$H}_{2}S{O}_{4$

$1.00kg{H}_{2}S{O}_{4}\times 0.98\times \frac{1mol\text{}{H}_{2}S{O}_{4}}{98.09\times {10}^{-3}kg{H}_{2}S{O}_{4}}=10.mol{H}_{2}S{O}_{4}$

$C{a}_{3}{\left(P{O}_{4}\right)}_{2}$

$1.00kgC{a}_{3}{\left(P{O}_{4}\right)}_{2}\times \frac{1mol\text{}C{a}_{3}{\left(P{O}_{4}\right)}_{2}}{310.18\times {10}^{-3}kgC{a}_{3}{\left(P{O}_{4}\right)}_{2}}$

$=3.2molC{a}_{3}{\left(P{O}_{4}\right)}_{2}\times \frac{3mol{H}_{2}S{O}_{4}}{1molC{a}_{3}{\left(P{O}_{4}\right)}_{2}}=9.6mol{H}_{2}S{O}_{4}$

Since$9.6mol<10mol$ ,

$C{a}_{3}{\left(P{O}_{4}\right)}_{2}$

is the limiting reactant.

Step 3

Finally, solve for the masses of the products:

$3.2molC{a}_{3}{\left(P{O}_{4}\right)}_{2}\times \frac{3molCaS{O}_{4}}{1molC{a}_{3}{\left(P{O}_{4}\right)}_{2}}\times \frac{135.15gCaS{O}_{4}}{1molCaS{O}_{4}}$

$=1.3\times {10}^{3}gCaS{O}_{4}$

Solve for the masses of the reactants:

Since

is the limiting reactant.

Step 3

Finally, solve for the masses of the products:

zurilomk4

Answered 2022-01-11
Author has **35** answers

First, balance the reaction.

The molar mass of calcium phosphate is

Therefore 1.0kg of calcium phosphate

The molar mass of the sulfuric acid is

The 1.0kg mixture is 98% sulfuric acid, which is 0.98kg or 980g. Dividing by the 98.09g/mol gives 9.99 moles.

From the balanced reaction, 1 mol calcium phosphate reacts with 3 mol sulfuric acid.

Therefore, 3.22 mol calcium phosphate will react with 3(3.22)=9.66mol sulfuric acid.

There are 9.99 mol sulfuric acid, so it is the excess reactant. We use the limiting reactant of calcium phosphate for the calculations.

From the reaction, 1 mol calcium phosphate forms 3 mol calcium sulfate and 2 mol phosphoric acid.

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